Stuck on algebra: solving for y when I have y^(-1/3)?

y^(-1/3) = 1/quberoot of y

set it equal to x

qube both sides you get

x^3 = 1/y

y=1/x^3

when you have a negative exponent, you have to drop it below a one in a division problem, like such:

1 / (y^1/3)

Now, when you have a fraction as an exponent, here is what each number means.

The numerator is the power in which you raise the variable. The denominator is the power of the root you want from the variable. so, something raise to the 1/3 means this.

3rd root of y^1, or simply the cube root of y.

That should be straight forward enough to finish from here.

suppose you have

y^(a/b) = something..

then, apply Ln to both sides to get

(a/b)Ln(y) = Ln(something)

obtaining,

Ln(y) = Ln(something)/(a/b)

then,remove Ln, by using inverse. in other words, y= otherside.

here is an example.

let y^(-2/3) = x

then.. apply Ln to both sides..

getting Ln( y^(-2/3) ) = Ln (x)

this moves the exponent to the front..

giving, (-2/3) Ln y = Ln x

now, dived both side by fraction part.. or multiply by reciprocal.

giving Ln y =-3/2 (Ln x)

remove Ln by its inverse...

see if that helps..this would give, e^(-3/2 Lnx)

which can be reduced...

if y^-(1/3)=x

then x^3=1/y

Cube everything, you'll get y = RHS^3.